/sci/ - Science & Math

[Anonymous No.4697325]

lim((ln(2x+3)/(3x+3))

for x->-1

solve this please

[Anonymous No.4697343]

>>4697325
1/3
Just integrate using l'hopitals rule.

[Anonymous No.4697344]

>>4697325
ln(2/3)

Now get out

[Anonymous No.4697346]

>>4697344
disregard this i didnt read the expression carefully enough

[Anonymous No.4697350]

You're all rong it's ln(3)/3

[Anonymous No.4697370]

>>4697325
2/3

<div class="math"> \ln{2x+3} = ln{1+ 2(x+1)} \sim_{x \rightarrow -1} 2(x+1) </div> so,
<div class="math"> \frac{\ln{2x+3}}{3(x+1)} \sim_{x \rightarrow -1} \frac{2(x+1)}{3(x+1)} = \frac{2}{3} </div> which means that
<div class="math"> \lim_{x \rightarrow -1} \frac{\ln{2x+3}}{3(x+1)} = \frac{2}{3} </div>

[Anonymous No.4697371]

>>4697370
<div class="math"> \ln(2x+3) = \ln(1+ 2(x+1)) \sim_{x \rightarrow -1} 2(x+1)</div>

[Anonymous No.4697716]

It's 2/3

3(x+1)

x->-1
x+1>0
y=x+1
x=y-1

ln(2y-2+3)/3y
ln(2y+1)/y 3
ln(2y+1/2y 3/2
1x3/2
3/2

[Anonymous No.4697858]

>>4697716
Learn to math